replace-regexp-in-string: How to prepend a string to the entire string matched?

Question

I am adding divider lines to certain sections of the reports generated by the binary ledger [ https://www.ledger-cli.org/ ] through the function ledger-do-report in the ledger-mode Lisp library [ https://github.com/ledger/ledger-mode]. To accomplish this goal, I have chosen replace-regexp-in-string. I wish to prepend the divider line followed by a new line; i.e., "======\n".

Q:  Is there a built-in way to prepend the additional string without the need to create an empty substring at the beginning of the REGEXP; i.e., \\(\\)?

In the following example, I have begun my REGEXP with ^\\(\\) so that I can replace that number 1 subsexpression with the new string. However, I have a feeling that there is a more correct approach to simply prepend a string without using an empty placeholder in the REGEXP.

NOTE: The ^[ is defined by Emacs as: old-name: ESCAPE; general-category: Cc (Other, Control); decomposition: (27) ('^[')

;;;           $ 98,477.47  ASSETS
;;;  =>
;;;
;;;           $ 98,477.47  ^[[34mASSETS^[[0m
;;;
(setq report (replace-regexp-in-string
               "^\\(\\)[\s]+$[\s]\\([0-9,.]+\\)[\s]+^[\\[34m\\(ASSETS\\)^[\\[0m$"
               (concat (make-string (- (window-width) 1) ?=)" \n")
               report nil nil 1))

Answer 1

If LITERAL is nil then you can use "\\&" in the replacement to represent the entire matched string, and so your replacement text could be "PREFIX\\&".

Refer to C-h f replace-match for the behaviour of LITERAL.

(And if you didn't want special characters in the replacement text to be processed, you would generally want LITERAL to be non-nil, for safety.)